∫ (x^2 + 4x^3 + 2) / (6x) dx
Understand the Problem
The question is asking to perform the integration of the function (x^2 + 4x^3 + 2) divided by (6x) with respect to x. This involves applying techniques of integration to solve the expression.
Answer
The integral evaluates to: $$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Answer for screen readers
The result of the integration is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
Steps to Solve
- Simplify the integrand To make integration easier, we can divide each term in the numerator by the denominator:
$$ \frac{x^2}{6x} + \frac{4x^3}{6x} + \frac{2}{6x} $$
This simplifies to:
$$ \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} $$
- Set up the integral Now we set up the integral:
$$ \int \left( \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} \right) dx $$
- Integrate each term separately Integrate each term in the expression:
- For $\frac{x}{6}$:
$$ \int \frac{x}{6} dx = \frac{1}{6} \cdot \frac{x^2}{2} = \frac{x^2}{12} $$
- For $\frac{2}{3}x^2$:
$$ \int \frac{2}{3}x^2 dx = \frac{2}{3} \cdot \frac{x^3}{3} = \frac{2x^3}{9} $$
- For $\frac{1}{3x}$:
$$ \int \frac{1}{3x} dx = \frac{1}{3} \ln |x| $$
- Combine the results Now, combine the results from the three integrals:
$$ \int \left( \frac{x}{6} + \frac{2}{3}x^2 + \frac{1}{3x} \right) dx = \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
where ( C ) is the constant of integration.
The result of the integration is:
$$ \frac{x^2}{12} + \frac{2x^3}{9} + \frac{1}{3} \ln |x| + C $$
More Information
This integral demonstrates how to integrate rational functions by breaking them down into simpler terms. The logarithmic rule in integration comes from the integral of ( \frac{1}{x} ), which is a fundamental concept in calculus.
Tips
- Forgetting to break down the integral into simpler terms before integrating.
- Not applying the constant of integration ( C ) after completing the integral.